Question

The number of stereoisomers possible for the compound \( \text{CH}_3\text{CHBrCHBrCH}_3 \) is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

For compounds with two identical chiral centers, always check for meso forms because symmetry reduces the total number of stereoisomers.

Answer 1

The Correct Option is B

Concept:
The number of stereoisomers depends on the number of chiral centers and the presence of symmetry. Maximum stereoisomers: \[ 2^n \] where \(n\) is the number of chiral centers. However, meso forms reduce the total number.

Step 1: Identify chiral centers.
Compound: \[ \text{CH}_3\text{CHBrCHBrCH}_3 \] Both middle carbons are attached to: \[ H,\ Br,\ CH_3,\ \text{remaining carbon chain} \] Thus both are chiral centers. \[ n=2 \] Maximum possible stereoisomers: \[ 2^2=4 \]

Step 2: Check for meso form.
The molecule is symmetrical. Configurations: \[ (R,R),\ (S,S),\ (R,S) \] The \( (R,S) \) form has an internal plane of symmetry. Thus it becomes a meso compound.

Step 3: Count distinct stereoisomers.
Distinct stereoisomers:
• One pair of enantiomers: \( (R,R) \) and \( (S,S) \)
• One meso form Total: \[ 3 \] Hence: \[ \boxed{3} \]